Inorganic and Applied Chemistry

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Inorganic and Applied Chemistry

Example 5- G:

pH in a buffer solution

A buffer solution consists of 0.5 M acetic acid CH 3 COOH (Ka value of 1.8 · 10-5 M) and 0.5 M sodium

acetate CH 3 COONa. This solution consists of a weak acid (acetic acid) ad its corresponding weak base

(acetate ion). As the amount of the weak acid and weak base are similar (it is sufficient if there just in the

same order of magnitude) we have a buffer system.

The initial concentrations of acetic acid and the acetate ion are equally large which is why pH in this case

is simply equal to the pKavalue of acetic acid according to the buffer equation:





log log 1 4. 74

3

3

a CHCOO pKa pKa

pH pK CHCOOH

We wish to illustrate the buffer system buffer effect by calculating what the pH of the solution becomes

when 0,01 moles of solid NaOH is added to 1.0 litre of the buffer solution. As NaOH is a strong base we

assume that all NaOH dissociate completely. The determining components in the solution is thereby:

CH 3 COOH , Na+, CH 3 COO-, OH- and H 2 O

As the solution contains a large amount of the strong base OH- the following reaction will proceed

primarily to the right:

CH 3 COOH (aq) + OH-(aq)  CH 3 COO-(aq) + H 2 O(l)

It is advantageous to divide the problem into two parts. First we assume that the reaction above runs

dominantly to the right and secondly; we will perform equilibrium calculations.

CH 3 COOH (aq) + OH-(aq)  CH 3 COO-(aq) + H 2 O(l)

Prior to

reaction:

1,0 L · 0,5 mol/L =

0,5 mol

0,01 mol 1,0 L · 0,5 mol/L =

0,5 mol

After reaction: 0,5 mol – 0,01 mol 0 mol 0,5 mol +0,01 mol =

0,51 mol

Now the problem may be treated as an equilibrium problem. The only difference from the procedure above

is that it is only some of the OH- ions that have reacted with CH 3 COOH. We will now look at initial and

equilibrium conditions analogously earlier examples. The initial concentrations are:

[CH 3 COOH] 0 = 0.49 M

[CH 3 COO-] 0 = 0.51 M

[H 3 O+] 0 * 0 (the autoprotolysis of water is neglected)

And the final concentrations are thereby:

[CH 3 COOH] = (0.49 – x) M

[CH 3 COO-] = (0.51 + x) M

[H 3 O+] = x M

Acids and bases